[最短路][期望DP]luogu P1850 换教室

https://www.luogu.org/problemnew/show/P1850

分析

先全部跑一边dij，把距离求出来

我们设f[i][j][0/1]表示第i个时段，用了j次换教室机会，当前教室有无使用机会的最小期望

$f[i+1][j][0]=min(f[i][j][0]+dis[c[i]][c[i+1]],f[i][j][1]+dis[c[i]][c[i+1]]*(1-p[i])+dis[d[i]][c[i+1]]*p[i])$

$f[i+1][j+1][1]=min(f[i][j][0]+dis[c[i]][c[i+1]*(1-p[i+1])+dis[c[i]][d[i+1]]*p[i+1],f[i][j][1]+dis[c[i]][c[i+1]]*(1-p[i])*(1-p[i])+dis[c[i]][d[i+1]]*(1-p[i])*p[i+1]+dis[d[i]][c[i+1]]*p[i]*(1-p[i+1])+dis[d[i]][d[i+1]]*p[i]*p[i+1]$

 

#include <iostream>
#include <cstdio>
#include <queue>
#include <memory.h>
using namespace std;
const int N=2e3+10;
struct Heap {
    int i,dis;
    friend bool operator < (Heap a,Heap b) {
        return a.dis>b.dis;
    }
};
struct Graph {
    int v,w,nx;
}g[2*90010];
int cnt,list[N];
int dis[N][N],c[N],d[N];
double f[N][N][2],p[N];
int n,m,v,e;

void Add(int u,int v,int w) {
    g[++cnt]=(Graph){v,w,list[u]};list[u]=cnt;
    g[++cnt]=(Graph){u,w,list[v]};list[v]=cnt;
}

void Dij(int v0) {
    priority_queue<Heap> q;
    while (!q.empty()) q.pop();
    q.push((Heap){v0,0});dis[v0][v0]=0;
    while (!q.empty()) {
        Heap a=q.top();q.pop();
        int u=a.i,d=a.dis;
        if (dis[v0][u]<d) continue;
        for (int i=list[u];i;i=g[i].nx)
            if (dis[v0][g[i].v]>dis[v0][u]+g[i].w)
                dis[v0][g[i].v]=dis[v0][u]+g[i].w,q.push((Heap){g[i].v,dis[v0][g[i].v]});
    }
}

int main() {
    scanf("%d%d%d%d",&n,&m,&v,&e);
    for (int i=1;i<=n;i++) scanf("%d",&c[i]);
    for (int i=1;i<=n;i++) scanf("%d",&d[i]);
    for (int i=1;i<=n;i++) scanf("%lf",&p[i]);
    for (int i=1,u,v,w;i<=e;i++) scanf("%d%d%d",&u,&v,&w),Add(u,v,w);
    memset(dis,0x3f,sizeof dis);
    for (int i=0;i<=v;i++) dis[0][i]=0;
    for (int i=1;i<=v;i++) Dij(i);
    for (int i=0;i<=n;i++)
        for (int j=0;j<=m;j++)
            for (int k=0;k<2;k++) f[i][j][k]=2147483647;
    f[0][0][0]=f[0][0][1]=0;
    for (int i=0;i<n;i++)
        for (int j=0;j<=min(i,m);j++) {
            f[i+1][j][0]=min(f[i+1][j][0],f[i][j][0]+dis[c[i]][c[i+1]]);
            f[i+1][j][0]=min(f[i+1][j][0],f[i][j][1]+dis[c[i]][c[i+1]]*(1.0-p[i])+dis[d[i]][c[i+1]]*p[i]);
            if (j<=min(i,m-1)) f[i+1][j+1][1]=min(f[i+1][j+1][1],f[i][j][0]+dis[c[i]][c[i+1]]*(1.0-p[i+1])+
            dis[c[i]][d[i+1]]*p[i+1]);
            if (j<=min(i,m-1)) f[i+1][j+1][1]=min(f[i+1][j+1][1],f[i][j][1]+dis[c[i]][c[i+1]]*(1.0-p[i])*(1.0-p[i+1])+
            dis[c[i]][d[i+1]]*(1.0-p[i])*p[i+1]+dis[d[i]][c[i+1]]*p[i]*(1.0-p[i+1])+dis[d[i]][d[i+1]]*p[i]*p[i+1]);
        }
    double ans=2147483647;
    for (int i=0;i<=m;i++)
        for (int j=0;j<=min(1,i);j++) ans=min(ans,f[n][i][j]);
    printf("%.2lf",ans);
}